2024 WAEC CHEMISTRY: (6474)

Number 1biii corrected and now authentic. NO.1biii corrected use this.

(1a)

Volume of burrete = 50.00cm³

Volume of pipette = 25.00cm³

Indicator = Methyl orange

TABULATE

Final burette readings (cm³); 37.50 | 22.50 | 37.50 | 44.50

Initial burette readings(cm³); 0.00 | 0.00 | 0.00 | 0.00

Volume f Acid used (cm³); 37.50 | 22.50 | 22.50 | 22.50

Volume of A used = 1st + 2nd + 3rd/3

= 22.50+22.50+22.50/3

Average volume of A used = 22.50cm³

(1bi)

Concentration of A ( CA )?

2.03g in 500cm³ of solution

Volume = 500cm³/1000 = 0.500dm³

In g/dm³ = 2.03/0.5 = 4.06g/dm³

To find concentration in mol/dm³

Conc. of A in mol/dm³ = Conc. of A in g/dm³/Molar mass

Conc. of A in mol/dm³ = 4.06/36.5

= 0.1112mol/dm³

Molar mass of A (HCL) = 1+35.5 = 36.5g/mol

(1bii)

Number of mole of the Acid in average titre

Na = CaVa

= 0.1112 × 22.50

= 2.52moles

(1biii)

Nb= ? Na= 3, Ca= 0.1112mol/dm³, Cb = 0.12mol/dm³ , Va = 22.50cm³, Vb = 25.00cm³

CaVa/CbVb= nA/nB

0.1112×22.50/0.12×25 = 3/nB

nB × 2.502 = 9

nB = 9/2.502 =

nB = 3.597moles

(1biv)

Mole ratio of Acid to base

nA:nB

= 3:3

= 1:1

============================

(2)

(2a)

Test; C + 10cm³ of distilled water + filter 

Observation; C dissolves partially to give a light green solution. Blue filtrate and green residue

Inference; C is a mixture of soluble and insoluble salt

(2bi)

Test; 2cm³ of filtrate + NaOH in drops in excess

Observation; Blue precipitate remains 

Inference; CU²+ is present

(2bii)

Test; 2bi + warm 

Observation; a colourless gas with a chocking smell which turns moist red litmus blue and forms dense white fumes with hydrogen chloride gas 

Inference; NH³ from NH⁴+

(2biii)

Test; 2cm³ of filtrate + BaCl² + excess HCl

Observation; white precipitate, precipitate Remains

Inference; SO²- CO3²- or SO4²- is present, SO4²- confirmed 

(2ci)

Test; Residue + HNOg

Observations; effervescence of a colourless and odourless gas which turns like water milky and turns moist blue to litmus red 

Inference; CO2 from CO3²-

(2cii)

Test; 2ci + NH3 in drops in excess 

Observations; blue precipitate. Precipitate dissolved to give a deep blue solution

Inference; CU2+ present 

CU2+ confirmed

===============================

(3)

(3ai) 

The value will increase 

(3aii)

The occur as a result of the decrease in the concentration of base due to the added volume of water 

(3bi) 

There will be no visible reaction because copper is less than Zinc in the electrochemical series

(3bii) 

It absorb water and become sticky because it is hygroscopic

(3biii)

The solution turns pink

(3c)

When NaOH is added to the solution of zn³+, a white precipitate is formed which later dissolve in excess NaOH due to the formation of zinc hydroxide

==============================

Completed 

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